About a type of limit

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  • #1
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Would this ##f'(x)>g'(x)\,\forall x\in [a,\infty)\text{ and }f,\,g\underset{\infty}{\to}0\Rightarrow \lim_{x\to\infty}f(x)/g(x)=0## hold if ##\frac{f(x)}{g(x)}\neq c##?
 

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  • #2
Math_QED
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It is unclear to me what you mean. Can you be a little more careful describe what you mean? You have an ##\implies## and then an 'if' after the implies.

What is ##a## ? What is ##c##? Where are ##f,g## defined?
 
  • #3
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Would this ##f'(x)>g'(x)\,\forall x\in [a,\infty)\text{ and }f,\,g\underset{\infty}{\to}0\Rightarrow \lim_{x\to\infty}f(x)/g(x)=0## hold if ##\frac{f(x)}{g(x)}\neq c##?
This is a little cryptic. Here's what I think you are saying:

"There is a theorem that if for some ##a \in \mathbb R, f'(x)>g'(x)\,\forall x\in [a,\infty)## and ##f,\,g\underset{\infty}{\to}0## (Does this mean ##\lim_{x\to\infty}f, g = 0##?) then ##\lim_{x\to\infty}f(x)/g(x)=0##. Does the theorem hold if ##\frac{f(x)}{g(x)}\neq c##?"

Or are you asking if there is such a theorem?
 
  • #4
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If I understood correctly, ##1/x ,1/(x+1)## are counters.
 
  • #5
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It is unclear to me what you mean. Can you be a little more careful describe what you mean? You have an ##\implies## and then an 'if' after the implies.

What is ##a## ? What is ##c##? Where are ##f,g## defined?
This is a little cryptic. Here's what I think you are saying:

"There is a theorem that if for some ##a \in \mathbb R, f'(x)>g'(x)\,\forall x\in [a,\infty)## and ##f,\,g\underset{\infty}{\to}0## (Does this mean ##\lim_{x\to\infty}f, g = 0##?) then ##\lim_{x\to\infty}f(x)/g(x)=0##. Does the theorem hold if ##\frac{f(x)}{g(x)}\neq c##?"

Or are you asking if there is such a theorem?
What I want to say is, if there're two functions ##f(x)## and ##g(x)## who are going to ##0## as ##x\to\infty##, and such that after a certain ##x\,(=a)## we have ##f'(x)>g'(x)## then ##\frac{f(x)}{g(x)} \underset{\infty}{\to} 0## provided that ##f(x)\neq cg(x)## where ##c\in\mathbb{R}##, elsewise the limit would equal ##c##.
I realised some holes in this through math.stackexchange. I was initially thinking of two decreasing functions, i.e both positive.
However, since what I want to say is if ##f## goes to ##0## faster than ##g## then ..., the correct way to translate this in math is ##|f'(x)|<|g'(x)|## (provided that both ##f## and ##g## are going to ##0##).
 
  • #6
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If I understood correctly, ##1/x ,1/(x+1)## are counters.
No, they don't come under this since they have the same rate of change.
 
  • #7
WWGD
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No, they don't come under this since they have the same rate of change.
Respective derivatives are ##-1/x^2 >-1/(x+1)^2##
 
  • #8
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Respective derivatives are ##-1/x^2 >-1/(x+1)^2##
Right sorry. I think, then, that I can say something about the convergence of the limit given the assumptions? And then it follows that it is ##0## if the functions are not equivalent at ##\infty##?
 
  • #9
PeroK
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What I want to say is, if there're two functions ##f(x)## and ##g(x)## who are going to ##0## as ##x\to\infty##, and such that after a certain ##x\,(=a)## we have ##f'(x)>g'(x)## then ##\frac{f(x)}{g(x)} \underset{\infty}{\to} 0## provided that ##f(x)\neq cg(x)## where ##c\in\mathbb{R}##, elsewise the limit would equal ##c##.
I realised some holes in this through math.stackexchange. I was initially thinking of two decreasing functions, i.e both positive.
However, since what I want to say is if ##f## goes to ##0## faster than ##g## then ..., the correct way to translate this in math is ##|f'(x)|<|g'(x)|## (provided that both ##f## and ##g## are going to ##0##).
I can't see any reason why this would be true.
 
  • #10
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I can't see any reason why this would be true.
$$N=n-m>0,\,\,\pm\frac{10^{-n}}{10^{-m}}=\pm10^{-n+m}=\pm10^{-N}=\pm\frac{1}{e^{N\ln10}}\underset{N\to \infty}{\to}0$$
By ##10^{k}## I just wanted to represent a kind of manual evaluation of a limit where the nominator is going to zero faster than the denominator. Maybe I should use ##O(10^{-n})##? Not sure, I need to study big-O notation.
 
  • #11
PeroK
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$$N=n-m>0,\,\,\pm\frac{10^{-n}}{10^{-m}}=\pm10^{-n+m}=\pm10^{-N}=\pm\frac{1}{e^{N\ln10}}\underset{N\to \infty}{\to}0$$
By $10^{k}$ I just wanted to represent a kind of manual evaluation of a limit where the nominator is going to zero faster than the denominator. Maybe I should use ##O(10^{-n})##? Not sure, I need to study big-O notation.
I can't see the relationship between that and what you posted before.
 
  • #12
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I can't see the relationship between that and what you posted before.
Well, if ##|f'(x)|<|g'(x)|## and they are both going to ##0##, it means that ##f## is going to ##0## faster, and so the limit should be ##0##.
 
  • #13
PeroK
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Well, if ##|f'(x)|<|g'(x)|## and they are both going to ##0##, it means that ##f## is going to ##0## faster, and so the limit should be ##0##.
That relationship means that ##g(x)## is changing more quickly than ##f(x)##.

You've got that the wrong way round.
 
  • #14
PeroK
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Well, if ##|f'(x)|<|g'(x)|## and they are both going to ##0##, it means that ##f## is going to ##0## faster, and so the limit should be ##0##.
In any case, you already have the counterexample: ##f(x) = \frac 1 2 g(x)##.

Even if you exclude that precise counterexample, you only need two functions where ##f(x) \approx \frac 1 2 g(x)##. For example:

##f(x) = (\frac 1 2 + \frac{\sin x}{x})g(x)##
 
  • #15
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That relationship means that ##g(x)## is changing more quickly than ##f(x)##.

You've got that the wrong way round.
If ##f## is going faster to ##0## than ##g##, then ##|df|## would be less than ##|dg|##, because it is already "at zero" so there isn't much to change.
In any case, you already have the counterexample: ##f(x) = \frac 1 2 g(x)##.

Even if you exclude that precise counterexample, you only need two functions where ##f(x) \approx \frac 1 2 g(x)##. For example:

##f(x) = (\frac 1 2 + \frac{\sin x}{x})g(x)##
Right, see post #8.
through the different counter-examples given, it is clear that this is not true. however, given the premises, it seems that the limit always converges (maybe), and in the special case that ##f\underset{\infty}{\not\sim}g##, it is zero (maybe). This is just an intuitive exercise, I just feel like this might work out.
 
  • #16
PeroK
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however, given the premises, it seems that the limit always converges (maybe), and in the special case that ##f\underset{\infty}{\not\sim}g##, it is zero (maybe). This is just an intuitive exercise, I just feel like this might work out.
Not that either. Just amend my last function:

##f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)##

Then the limit ##f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}## is undefined.
 
  • #17
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Not that either. Just amend my last function:

##f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)##

Then the limit ##f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}## is undefined.
wops
 
  • #18
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Not that either. Just amend my last function:

##f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)##

Then the limit ##f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}## is undefined.
It doesn't satisfy ##|f'(x)|<|g'(x)|## so it doesn't come under this.
 
  • #19
PeroK
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wops
Seriously, you ought to think about how I created these counterexamples. They only really involved some elementary geometrical thinking.

The only one we don't have is where ##\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = +\infty##.

Any ideas?
 
  • #20
PeroK
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It doesn't satisfy ##|f'(x)|<|g'(x)|## so it doesn't come under this.
With ##g(x) = e^{-x}## it all works out.
 
  • #21
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Seriously, you ought to think about how I created these counterexamples. They only really involved some elementary geometrical thinking.

The only one we don't have is where ##\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = +\infty##.

Any ideas?
you mean given the premises?
 
  • #22
PeroK
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  • #23
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you mean given the premises?
Maybe this is the result that could be proved? If ##|f'(x)| < |g'(x)|##, then ##f(x)## is changing less than ##g(x)##, so ##f(x)/g(x)## cannot become arbitrarily large. So:

##\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} \ne +\infty##?
 
  • #24
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Maybe this is the result that could be proved? If ##|f'(x)| < |g'(x)|##, then ##f(x)## is changing less than ##g(x)##, so ##f(x)/g(x)## cannot become arbitrarily large. So:

##\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} \ne +\infty##?
I thought you meant that there was an example where that limit goes to infinity...
But yes, it seems reasonable that given those premises the limit can't be arbitrarily large.
 
  • #25
PeroK
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I thought you meant that there was an example where that limit goes to infinity...
... looking for counterexamples is good way to see what might be true.
 

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